Question

If the equation of a conic section is `(x+6)^2/81-(y+1)^2/16=1`, how has its center been translated?

Answer 1

This is the equation of an hyperbola centered in the point `C(-6,-1)`, with semi-axes: `a=9` and `b=4`.

The equation of an hyperbola centered in `O(0,0)` with semi-axes `a` and `b` is:

`x^2/a^2-y^2/b^2=+-1`.

If the second member is `1` then the hyperbola has the two branches on the "left" and on the "right".

E.G.: `x^2/4-y^2/9=1`

graph{x^2/4-y^2/9=1 [-10, 10, -5, 5]}

If the second member is `-1` then the hyperbola has the two branches "up and "down":

E.G.: `x^2/4-y^2/9=-1`

graph{x^2/4-y^2/9=-1 [-10, 10, -5, 5]}

The equation of an hyperbola centered in `C(x_c,y_c)` with semi-axes `a` and `b` is:

`(x-x_c)^2/a^2-(y-y_c)^2/b^2=+-1`.

So, our hyperbola is:

`(x+6)^2/81-(y+1)^2/16=1`.

graph{(x+6)^2/81-(y+1)^2/16=1 [-20, 10, -20, 20]}